Solutions to Problem Set 8
1. water saturated with H2 and O2 gases at a total pressure of 5 atm with (2:1 = H2:O2)
Equation
5 atm = pH2+pO2 ;
Basis for the equation
pH2/pO2 = 2/1
Given
Eq. # 1
pH2 = (2/3) 5 atm = 3.333 atm
pO2 = (1/3) 5 atm = 1.666 atm
pH2 = KH(H2) xH2
3.333 atm = 7.76 ×104 atm xH2
xH2 = 3.333/7.76×104 = 4.295×10-5
pO2 = KH(O2) xO2
1.666 atm = 4.58 ×104 atm xO2
xO2 = 1.666/4.58×104 = 3.639×10-5
mole ratio of H2:O2 in solution was
4.295×10-5/3.639×10-5 = 1.18
or 54.1 mole% H2 45.9 mole% O2 in
the gases driven off
Answer
Henry’s law applied to H2 gas solubility
Concentration in solution is directly proportional to
the partial pressure of the gas above the solution
pH2 = partial pressure of H2 in the overhead gas
xH2 = mole fraction of H2 in the aqueous solution
KH (in atm) is given in provided table
Solubility of these gases do not strictly follow
Henry’s law, since the apparent Henry’s law
constant from experiment varies with overhead
partial pressure! Use the KH at the closest partial
pressure (as we have done here). Better yet,
make a least squares fitting of apparent KH vs p
and use the equation to find the appropriate KH to
use at the partial pressures of this problem.
2
3
4
2. vapor pressure lowering by a solute
Equation
Basis for the equation
pBz /p*Bz = xBz
74.01/74.66 = xBz = 0.9913
100 g C6H6/78 g mol-1 = 1.282 mol Bz
1.282/[1.282 mol + n] = 0.9913
n = 0.01125 mol solute
2 g/MW g mol-1 = 0.01125
MW = 178
178 = c 12 + h 1
c 12/178 = 0.944
c = 14
178 = 14(12) + h(1)
h = 10
The molecular formula is C14H10
Answer
Assuming ideal solution for the solvent, Raoult’s
law holds
Find n number of moles of solute in 100 g Bz
Find molecular weight of solute
For molecular formula CcHh
weight fraction of carbon in this formula is 94.4%
Solve for c
Substitute c into Eq 6
Eq. # 1
2
3
4
5
6
7
8
3. Freezing point depression by a solute
H2O(liq) ⇔ H2O(s)
Equation
Basis for the equation
μ* H2O,liquid(T,p*) = μ*H2O,vap (T,p*)
Pure liquid H2O in equilibrium with its vapor, *
means pure
For the vapor at a given temperature, behaving
like an ideal gas
Recall that we used dG = Vdp when dT = 0 and
replaced V by RT/p to get this.
If non-ideal vapor use RT ln (f/1atm) instead
H2O in aqueous solution in equilibrium with its
vapor. As above, except that partial pressure pH2O
of H2O in the vapor in equilibrium with liquid
solution is less than p*
μ*H2O,vap (T,p*) = μ\H2O,vap (T)
+ RT ln (p*/1atm)
μ H2O,liquid solution (T,p) = μH2O,vap (T,p)
μH2O,vap (T,p) = μ\H2O,vap (T)
+ RT ln (pH2O/1atm)
μ
- μ* H2O,liquid(T,p*)
= RT ln(pH2O/p*)
Eq. # 1
2
3
4
Eq. (3+4) minus Eq (1+2)
5
pH2O/p* = xH2O
μH2O,liquid solution (T,p) - μ* H2O,liquid(T,p*)
= RT lnxH2O
Assume Raoult’s law holds for the liquid solution
Substituting Eq 6 into Eq 5
6
7
For H2O(liq) ⇔ H2O(s) at equilibrium
at T*f and 1 atm,
μ*H2O,solid(T*f,1 atm)
= μ* H2O,liquid(T*f,1 atm)
For H2O(liq solution) ⇔ H2O(s) at
equilibrium at T and 1 atm,
μ*H2O,solid(T ,1 atm)
= μH2O,liquid solution(T ,1 atm)
μ*H2O,solid(T ,p) = μ* H2O,liquid(T,p)
+ RTlnxH2O
lnxH2O =
[μ*H2O,solid(T ,p) - μ*H2O,liquid(T,p)]/RT
lnxH2O = - [ΔfusG]/RT
Pure liquid water in equilibrium with ice at the
normal freezing point T*f
8
Aqueous solution in equilibrium with pure ice at
the new freezing point T
9
Applying Eq 10 and Eq 7
10
Rearrange Eq 11
11
H2O,liquid solution (T,p)
(∂lnxH2O /∂xH2O) = - (1/R)•
[ ∂(ΔfusG/T)/∂T]p•(∂T/∂xH2O)p
[ ∂(ΔfusG/T)/∂T]p = - ΔfusH/T2
(∂lnxH2O /∂xH2O) =[ΔfusH/RT2]•
(∂T/∂xH2O)p
12
To discover how new freezing point T depends on
the water concentration in the liquid solution xH2O,
p being constant, we need to evaluate (∂T/∂xH2O)p
Apply (∂ /∂xH2O) to Eq 13
Gibbs-Helmholtz equation which we derived
previously
Substitute Eq 15 into Eq 14
13
14
15
1/ xH2O = [ΔfusH/RT2]• (∂T/∂xH2O)p
dxH2O/ xH2O = (ΔfusH/R) dT/T2
∫ dxH2O/ xH2O = ∫ (ΔfusH/R) dT/T2
∫1xH2OdxH2O/ xH2O
= ∫T*fT (ΔfusH/R) dT/T2
ln xH2O = (ΔfusH/R){ - T -1 + T*f-1}
ln xH2O = (80/1.9872){ - 263-1 + 273-1}
= - 0.00561
xH2O = 0.9944
nH2O = 4000/18 = 222.22
xH2O = nH2O/(nH2O+nsolute)
0.9944 = 222.22/(222.22 +nsolute)
n solute = 1.25 mol
g CH3OH = 1.25(32) = 40
g C2H4(OH)2 = 1.25(62) = 77.5
Answers
Integrating
Between the limits of 1 mole fraction (pure liquid
water) and xH2O, and corresponding temperatures
T*f and new freezing temperature T.
Assume ΔfusH = 80 cal g-1 and is only mildly
dependent on T over this tiny temperature range
For T = 263 K
Using density of water = 1 g mL-1, given 4 L H2O
weighs 4000 g
Weights of methanol and ethylene glycol to be
dissolved in 4 L water to prevent ice formation at
263 K
16
17
18
19
20
21
22
23
4. Freezing point depression by a solute and osmotic pressure
Equation
Basis for the equation
ln xH2O = (ΔfusH/R){ - T -1 + T*f -1}
ln xH2O =(80/1.9872){ -272.74-1 +273-1}
= - 2.016×10-4
Vπ = - RT ln xH2O
Vπ = - RT(- 2.016×10-4)
π = - (8.31451Pa m3K-1mol-1) 298
•(- 2.016×10-4)/ 18×10-6 m3
π = 27750 Pa Answer
Following same derivation as in problem 3 above
Substituting ΔfusH = 80 cal g-1 and T*f = 273.15 for
water and given t = -0.406°C = 272.74 K
following derivation in Problem Set 7 prob. 1(b)
Substituting Eq 2 into Eq 3
Using the molar volume of water = 18×10-6 m3
based on molar mass of 18g, density of 1 g cm-3
at 298 K.
This answer is inaccurate because we assumed
that NaCl is undissociated in solution and the
solution is an ideal solution. To do the problem
correctly, we will need to use activities of the +
and - ions in the solution, all of which contribute
to the osmotic pressure. We will find out more
about ionic activities in Lecture notes part 9.
Use the sum of mole fractions = 1 and expand the
ln in a series
Alternatively, we also could have done
this.
ln xH2O = ln(1-xs) ≈ -xs -(1/2)xs2
- (1/3)xs3 - ...
Vπ = RT xs
The relation between molar volume of liquid water
the osmotic pressure and the mole fraction of salt
Eq. # 1
2
3
4
5
4a
5a
5. 10 L bulbs, T = 298 K throughout
Vapor pressure of water = 24 mm Hg = 0.0316 atm
solubility of gas in water is 2 moles per 1000 g water
Equation
2 atm•10 L =
ngas (0.0820578 L atm K-1 mol-1) 298 K
ngas = 0.818 mol
mol of gas that went into 100 g liquid
water is 0.20 mol
Gas phase therefore has 0.818-0.20 =
0.618 mol gas in 19.90 L
plus that amount of water vapor that
corresponds to a partial pressure of
0.0316 atm and 19.90 L
Basis for the equation
Eq. # 1
Initially with stopcock closed, the left bulb has
water and water vapor. Volume of liquid water is
very close to100 mL. Available volume for water
vapor is (10. – 0.100) = 19.90 L.
If behaving as an ideal gas
Right bulb only has 0.818 mol gas in the 10 L
volume.
At equilibrium after stopcock is opened, some gas 1
goes into the left bulb into solution, liquid water
remains in the left bulb and the gas phase now
occupies a total of 19.90 L containing water vapor
plus gas
First assume amount of liquid water that has
1
turned into water vapor is much much less than
100 g. then later test whether this is true or not.
Using the given gas solubility
0.0316 atm•19.90 L
= n (0.0820578 L atm K-1 mol-1) 298 K
n water vapor = 0.0257 mol
g water vapor = 0.0257mol(18 g mol-1)
= 0.46 g
Assuming water vapor behaves as an ideal gas
The mass of liquid water that this corresponds to
is 0.46 g, occupying a volume of 0.46 mL
So our assumption that the amount of liquid water
that has turned into water vapor is much, much
less than 100 g is quite good since 0.46 << 100.0.
Also, the volume available to the gas phase is
19.90046, that is, 19.90 is therefore close enough.
2
In summary, equilibrium condition has
99.54 g of liquid water in the left bulb,
and in the gas phase spread into both
bulbs are 0.0257 mol water vapor and
0.618 mol of the gas. The partial
pressure of the gas is
0.618 mol(0.0820578 L atm K-1 mol-1)
298 K/ 19.9 L = 0.759 atm
The total pressure in the gas phase is
therefore 0.759 + 0.0316 = 0.791 atm
Answer
6. At 40°C 1 g of CO2 will dissolve under 1 atm pressure in 1000 g H2O.
At 20°C 1.7 g of CO2 will dissolve under 1 atm pressure in 1000 g H2O.
Equation
Basis for the equation
At 40°C 1 g of CO2 will dissolve under
1 atm pressure in 1000 g H2O.
At 40°C 2 g of CO2 will dissolve under
2 atm pressure in 1000 g H2O.
Thus, we must have no more than 2 g
of CO2 in 1000 g H2O at 20°C
At 20°C 1.7 g of CO2 will dissolve
under 1 atm pressure in 1000 g H2O.
this corresponds to a pressure of
(2/1.7) = 1.176 atm of CO2 at 20°C
Answer
1.7 g is 1.7/44 = 0.0386 mol CO2
1000 g is 1000/18 = 55.555 mol H2O
1
Eq. # Given
We are allowed to assume Henry’s law holds
pCO2 = KH(CO2) xCO2 at 40°C
Using the given 2 atm pressure for the bursting
limit (and we are neglecting the vapor pressure of
water so that total pressure is entirely due to CO2)
Given
Using pCO2 = KH(CO2) xCO2 at 20°C
Using molar masses of CO2 and H2O
1 atm
Find the KH(CO2) at 20°C
= KH(CO2) 0.0386/[55.555 + 0.0386] Applying pCO2 = KH(CO2) xCO2 at 20°C
= KH(CO2) 0.000695
KH(CO2) = 1439.3 atm
1
2
3
1.176 atm of CO2 at 20°C therefore
corresponds to
xCO2 = 1.176/1439 = 0.00082 Answer
Substituting the value of KH into
pCO2 /KH(CO2) = xCO2 at 20°C
7. T vs X diagram for O2 - N2 mixture at 1 atm.
Let L be the amount of liquid remaining in the still at a given instant during the
distillation. After distilling a differential amount dL, the amount of liquid will be (L - dL)
and the amount of vapor formed will be dL. At the instant, the composition of the liquid
is XO2 and the composition of the vapor is YO2. Consider material balance of O2
component: O2 overall = O2 in the remaining liquid + O2 in the vapor
(L)(XO2) = (L-dL)(XO2-dXO2) + (dL)(YO2+dYO2)
Neglecting products of infinitesimals,
(L)(XO2) = (L)(XO2) - (XO2)(dL) -(L)(dXO2) + (dL)(YO2)
or (YO2 - XO2)(dL) = (L)(dXO2)
(dL) = (dXO2)
(L)
(YO2 - XO2)
LIQUID: the overall composition is 0.20 O2 and 90 % is distilled, that is 10 % remaining
Integrating, from 10% up to 100%, ln (L2/L1) = ln (1.00/0.10) = 2.303 = ⌠ 0.20 (dXO2)
⌡X (YO2 - XO2)
-1
from molefraction 0.20 O2 to XO2. Plot the curve (YO2 - XO2) vs XO2 (see figure above
right). Now find the value of XO2 at which the integral underneath the curve between
XO2 and 0.20 is equal to 2.303.
We read XO2 = 0.23 is the composition of the remaining liquid. Answer
And (reading T for which XO2 = 0.23 in the figure above left) this occurs at T = 79.1 K.
Answer
Calculate activities: aO2 = pO2/pO2* YO2 = pO2/ptot ptot =1.0 atm aO2 = YO2•(1atm/pO2*T)
We can get p*T using the Clausius Clapeyron Eq
XO2
aO2
0.0
0
0.081
0.022
/pO2*78K
0.216
0.334
0.522
0.662
0.778
0.885
0.068
/pO2*79K
0.12
/pO2*80K
0.236
/pO2*82K
0.369
/pO2*84K
0.522
/pO2*86K
0.696
/pO2*88K
1.0
1.0
8. T vs X diagram for CCl4 - C2Cl4 mixture at 1 atm.
(a) LIQUID: the overall composition is 0.30 CCl4 and 50 % is distilled, that is 50 %
remaining
Integrating, from 50% up to 100%, ln (L2/L1) = ln (1.00/0.50) = 0.693 = ⌠ 0.30 (dX )
⌡X (Y - X )
from molefraction 0.30 CCl4 to X . Plot the curve (Y - X )-1 vs X (see figure above right).
Now find the value of X at which the integral underneath the curve between X and 0.30
(hatched area a) is equal to 0.693.
We read X = 0.044 is the CCl4 composition of the remaining liquid. To find the Y (in the
collected distillate), apply mass balance for CCl4:
CCl4 overall = CCl4 in the remaining liquid + CCl4 in the collected distillate
0.30 L = 0.044(L/2) + Y(L/2); Solve for Y, Y = 0.556
Answer
(b) LIQUID: the overall composition is 0.50 CCl4 and is distilled until remaining liquid
is 0.20 CCl4 . If the distillate is removed continuously, then at the instant that the liquid is
0.20 CCl4, the vapor coming off is 0.67 CCl4 (as seen in the table).
Answer
On the other hand, if looking for Y composition in the total collected distillate, first find
the area underneath the curve (figure above right) between X = 0.20 and X = 0.50, and
ln (1.00/f) = ⌠ 0.50 (dX )
⌡0.20 (Y - X )
The hatched area b provides the value of the fraction f of the liquid remaining.
We find the area = 0.93 = ln (1/f); 1/f = 2.53; thus f = 0.395
Mass balance of CCl4 is applied, knowing f = 0.395, to find Y the CCl4 composition of
the total collected distillate.
0.50 L = 0.20L(0.395) + Y L(0.605) Solve for Y, Y = 0.70
Answer
This is larger than 0.67. Since CCl4 is the lower boiling component, at the beginning
where the liquid is 0.50 CCl4, the first vapor to come off is about 0.90 CCl4, and this Y
value diminishes to the point where the remaining liquid is 0.20 CCl4 the vapor coming
off is 0.67 CCl4, as we saw in the table. Therefore the total collected vapor has to have
Y CCl4 greater than 0.67
9. 1 g of antibiotic substance MW = 10000 g mol-1 is dissolved in 100 g H2O.
Equation
Basis for the equation
Eq. # 1 g is 1 /10000 = 0.0001 mol
substance
100 g is 100/18 = 5.5555 mol H2O
xs = 0.0001/(5.5555+.0001) = 1.8×10-5
1-xs = xH2O
ln xH2O = ln(1-xs) ≈ -xs -(1/2)xs2
-(1/3)xs3 - ...
ln xH2O = -xs is close enough
pA /p*A = xA
1
Expansion of ln(1-xs) for xs <<< 1, as in this case
Raoult’s law holds for the solution, here solvent A
is H2O
2
3
4
Since xs is very small here, xH2O is
very close to 1 so pA will be quite
close to p*A thus, this is not a practical
test.
freezing point depression following derivation in
problem 3 above
-xs = (80/1.9872) (T-T*)/T*2
-1.8×10-5 = 40.26[ - (T*-T)/2732]
(T*-T) = 0.033 deg
Answer
freezing point depression is small, not
a very practical test
6
boiling point elevation, by an analogous derivation
-xs = (539.43/1.9872) (T*-T)/3732
-1.8×10-5 = 271.45[ - (T*-T)/3732]
(T-T*) = 0.0092 deg
Answer
boiling point elevation is very small,
not a very practical test
Vπ = RT xs
π = RT xs/V
π = (8.31451Pa m3K-1mol-1) 298
•(1.8×10-5 )/ 18×10-6 m3
= 2477 Pa or 18.57 mm Hg Answer
This is easily measurable and this is
the preferred test, especially since
only a few mg of substance is
available for testing.
Answer
5
7
8
osmotic pressure following derivation in Problem
Set 7 prob. 1(b) and problem 4 above
9
Using the molar volume of water= 18×10-6 m3
10
10. vapor pressure of zinc at 773 K, Tboil = 1180 K
Equation
Basis for the equation
ln(p/1atm)=(Hvap-Hliq) /RTboil]
•[1-Tboil/T]
If we know ΔvapH at Tboil ,we can find
vapor pressure p at temperature T
ΔvapH for the phase change at 1180 K
is 21 (1180) = 24780 cal mol-1
For any temperature T, we need to
correct this ΔvapH value by using
(∂H/∂T)p dT integrated from 1180 K to
T for both the liquid and the vapor
ΔvapH T = ΔvapH 1180 K
+ ∫1180T [Cp(vap) - Cp(liq)] dT
ΔvapH T = ΔvapH 1180 K
+ ∫1180T [4.968 - 7.09
+ 1.15x10-3 T] dT
ΔvapH T = ΔvapH 1180 K
+ 2.122( T-1180)
+ 1.15x10-3 (1/2)[T2 -11802]
ΔvapH 773 = ΔvapH 1180 K
- 2.122(773-1180)
+ 0.000575 [7732-11802]
= ΔvapH 1180 K + 863.6 – 457.0
= ΔvapH 1180 K + 406.5
ΔvapH 773 = 24780 + 406.5 = 25186
As expected, ΔvapH is higher at the
lower temperature.
Our answer would be in error if we had
assumed ΔvapH is independent.of T
ln(p/1atm)= (ΔvapH 773/R)•
[1- (1180/773)]/1180
ln(p/1atm) = 12672• -4.46×10-4 = -5.65
p/1 atm = exp(-5.65) = 0.00352
Answer
On the other hand, if we used
ΔvapH 1180 K = 24780
ln(p/1atm) = 12468• -4.46×10-4 = -5.56
p/1 atm = exp(-5.56) = 0.00385
The vapor pressure would be 9% too
high if we assume that ΔvapH is
independent.of T
Answer
As we derived in Problem Set 7 prob 6 when we
could assume that the vapor behaved like an ideal
gas. (the Clausius Clapeyron equation)
Since this value is not given, we estimate the
value of ΔvapH at Tboil
Trouton’s rule: ΔvapH /Tboil ≈ 21 cal mol-1 K-1
Eq. # 1
2
3
Here we are allowed to assume that Cp (vap) =
(5/2)R = 4.968 cal mol-1
and we are given
Cp(liq) = 7.09 – 1.15x10-3 T in cal mol-1
Using Eq 1 (the Clausius Clapeyron equation)
Actually our largest error is in assuming that
Trouton’s rule holds.
4